Given :
Equation of plane : \(2 x+4 y-z=1\)
Equation of line
$$
\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}
$$
To Find : Point of intersection of line and plane.
Let \(P(a, b, c)\) be point of intersection of plane and line.
As point \(P\) lies on the line, we can write,
$$
\Rightarrow a=2 k+1, b=-3 k+2, c=4 k-3 \ldots \ldots \ldots(1)
$$
Also point \(P\) lies on the plane
$$
\begin{aligned}
&2 a+4 b-c=1 \\
&\Rightarrow 2(2 k+1)+4(-3 k+2)-(4 k-3)=1 \ldots . . f r o m(1) \\
&\Rightarrow 4 k+2-12 k+8-4 k+3=1 \\
&\Rightarrow-12 k=-12
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \mathrm{k}=1 \\
&\therefore \mathrm{a}=2(1)+1=3 \\
&\mathrm{~b}=-3(1)+2=-1 \\
&\mathrm{c}=4(1)-3=1
\end{aligned}
$$
Therefore, co-ordinates of point of intersection of given line and plane are
P = (3,-1,1)
